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Question
Find the angle between the vectors \[\vec{a} = \hat{i} - \hat{j} + \hat{k} \text{ and } \vec{b} = \hat{i} + \hat{j} - \hat{k} .\]
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Solution
\[\text{ We have }\]
\[ \vec{a} = \hat{i} - \hat{j} + \hat{k} \text{ and } \vec{b} = i + j - k \]
\[\text{ Let }\theta\text{ be the angle between } \vec{a} \text{ and } \vec{b} . \]
\[\left| \vec{a} \right| = \sqrt{\left( 1 \right)^2 + \left( - 1 \right)^2 + \left( 1 \right)^2} = \sqrt{3}\]
\[\left| \vec{b} \right| = \sqrt{\left( 1 \right)^2 + \left( 1 \right)^2 + \left( - 1 \right)^2} = \sqrt{3}\]
\[\text{ and }\]
\[ \vec{a} . \vec{b} = 1 - 1 - 1 = - 1\]
\[\text{ Now }, \]
\[\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|} = \frac{- 1}{\sqrt{3}\sqrt{3}} = \frac{- 1}{3}\]
\[ \therefore \theta = \cos^{- 1} \left( \frac{- 1}{3} \right)\]
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